If $x$ and $y$ are two vectors such that $\|y\|_2=1$ and $x$ is in the non-negative orthant then how to show that the condition $$x^Ty\leq1$$ leads to the condition that $$\|x\|_2\leq1.$$ I know that Cauchy Schwarz Inequality means that $$\|x^Ty\|_2\leq \|x\|_2\|y\|_2$$ but I still do not understand how in this case $$x^Ty=\|x^Ty\|_2.$$ Any help in this regard will be much appreciated. Thanks in advance.

$\begingroup$Any vector $x$ orthogonal to $y$ trivially satisfies $x^\top y = 0 \leq 1$ but $\| x \|$ can be as large as you want. However, if you want $x^\top y \leq 1$ for all $y$ such that $\| y \| = 1$, it is straightforward.$\endgroup$
– VHarisopMay 9 '18 at 0:31

$\begingroup$@VHarisop yes I want it for all $y$. How to show that? What straight forward thing I am missing here?$\endgroup$
– Frank MosesMay 9 '18 at 0:37

$\begingroup$Because this is the "most pathological" example you can find in the unit sphere. After you've solved this example, all the others follow from Cauchy-Schwarz.$\endgroup$
– VHarisopMay 9 '18 at 0:52

$\begingroup$Is it right to say that (I think this was what you meant from pathalogical case) "we should take $y=\frac{x}{\|x\|}$ because for any other $y$ of the form $y=\frac{p}{\|p\|}$ will always produce a smaller value than for $y=\frac{x}{\|x\|}$ case due to Cauchy Schwarz Inequality, which becomes equality only when $y$ is a linear combination of $x$"?$\endgroup$
– Frank MosesMay 9 '18 at 1:09

1

$\begingroup$@FrankMoses: Yes, this is essentially what you are doing by considering $y$ collinear to $x$.$\endgroup$
– VHarisopMay 9 '18 at 1:30